What name shall I give to this problem? - 2

Construct $DI$ and $EJ$ parallel to $BF$ with $I$ and $J$ on $AC$ .

Since, $BD = DE = EC$ . It implies that $FI = IJ = JC$ . So, $FJ : FC = 2 : 3$ . But since $AF = FC$ . So, $AF : FJ = 3 : 2$ .

Observe that, $AQ : QE = AF : FJ = 3:2$ .

Now let us do another construction. Draw a line parallel to $AD$ passing through $Q$ and meeting $BC$ at $Z$ .

Observe that $DZ : ZE = AQ : QE = 3 : 2$ .

So, $BD : DZ = DE : DZ = DZ + ZE :DZ = \dfrac{2}{3} +1 = 5 : 3$ .

But, it is clear that $BP : PQ = BD : DZ = 5 :3$ .

So, $a=5$ and $b=3$ . Therefore, $10a+b = \boxed{53}$ .

By Menelaus Theorem of Transversal Line, we need to find the ratio of $\frac{EA}{AQ}$ to find the ratio of $\frac{BP}{PQ}$

Thus, we have the $\frac{EA}{AQ}$ via the transversality of point $B, Q, F$ $\frac{CB}{BE} \frac{EQ}{QA} \frac{AF}{FC} = 1$ $\frac{3}{2} \frac{EQ}{QA} \frac{1}{1} = 1$ $\frac{EQ}{QA} = \frac{2}{3}$

In conclusion, we have $\frac{EA}{AQ} = \frac{5}{3}$

Then, by finding the ratio of $\frac{BP}{PQ}$ , we use the transversality of $A, P, D$ $\frac{EA}{AQ} \frac{QP}{PB} \frac{BD}{DE} = 1$ $\frac{5}{3} \frac{QP}{PB} \frac{1}{1} = 1$ $\frac{QP}{PB} = \frac{3}{5}$ $\frac{BP}{PQ} = \frac{5}{3}$

$a = 5, b = 3 \Rightarrow 10a + b = 53$

Let be on line segment such that is parallel to

Let , so therefore

Since is parallel to , we know that so that means

That also leads to more similiar triangles which means

Notice that , which means

Now let which means and

Also we notice . Because , is the midpoint of .

That means which leaves

and so

Contruction $\begin{aligned}1.~&\text{Join} FE \\ 2.~& \text{Draw }SR~\| ~BC\end{aligned}$

Let,

$\begin{aligned}PD&=x\\EF &=y\\AP &= z\end{aligned}$

Since

$\begin{aligned}BD &= DE, \\ y &= 2x\\ x &= \frac{y}{2}\end{aligned}$

$\|\text{ly}, \\\begin{aligned}z+x &= 2y\\ \Rightarrow z&= \frac{3}{2}y\\ \Rightarrow \dfrac{FQ}{QP} &= \dfrac{y}{z} = \frac{2}{3}\end{aligned}$

Now,

$\begin{aligned} BP &= PQ + QF\\ \Rightarrow BP &= \left(\frac{2}{3} + 1\right) PQ\\ \Rightarrow \dfrac{a}{b} &= \dfrac{BP}{PQ} = \dfrac{5}{3}\\ \Rightarrow a = 5 ~~&~~b = 3\end{aligned}$

$\therefore \boxed{10a + b = 53}$